y^'(1+e^x)=e^{x-y}

Lösung

y^{^{'}} (1 + e^{x}) = e^{x - y}

y = ln (e^{x} + c_{1})

Schritte zur Lösung

y^{'} (x) = e^{x - y}

Solve separable ODE:

y = ln (e^{x} + c_{1})

y = ln (e^{x} + c_{1})

\frac{\partial}{\partial x} (sin (x - y) + (x + y))

\frac{\partial}{\partial x} (4 x + 2 y)

t an g e n t f (x) = 5 tan (x), (\frac{π}{4}, 5)

\frac{d}{d x} (sec (1 + x^{2}))

\int \frac{63}{1 - cos ( 7 x )} d x